In a recent report, Joe's, a Memphis-style barbecue chain, states that 11% of its customers order for delivery. A random sample of 6 Joe's customers is chosen. Find the probability that at most 1 of them order for delivery. " Do not round your intermediate computations, and round your answer to three decimal places.

If there were four girls and four boys to choose from, the new sample size is one that will be a total of 16 possible combinations.

When making selections for the student council's representatives, the sample size belongs to the whole of the feasible options.

To know the sample size, we must reason every conceivable permutation involving a single female and a single male.

So, if there were four girls and four boys to choose from, the new sample size would be:

There are:

4 choices for girls

4 choices for boys.

So, for each girl's choice, there would be 4 corresponding choices for boys.

Hence it will be:

4 x  4 = 16

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If a large airline wants to estimate its average number of unoccupied seats per flight over the past year Then a 92% confidence interval for the population mean number of unoccupied seats per flight is (10.334, 12.266) seats.

To construct a confidence interval for the population mean number of unoccupied seats per flight, we will use the following formula CI = X ± z*(σ/√n)

Where:

X = sample mean = 11.3 seats

σ = sample standard deviation = 4.3 seats

n = sample size = 300

z = z-score corresponding to the desired confidence level of 92%, which we can look up in a standard normal distribution table or use a calculator. For a 92% confidence level, the z-score is 1.75.

Plugging in the values, we get:

CI = 11.3 ± 1.75*(4.3/√300)

CI = 11.3 ± 0.966

Therefore, the 92% confidence interval for the population mean number of unoccupied seats per flight is (10.334, 12.266) seats. This means we can be 92% confident that the true population means a number of unoccupied seats per flight falls within this interval.

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The  expected value and standard error for the number of winning plays is $3.1 and $ 2.21.

The population mean's likelihood to differ from a sample mean is indicated by the standard error of a mean, and simply standard error.

It reveals how much what the sample mean will change if a study were to be repeated with fresh samples drawn from a single population.

Chance of winning $3 = 2/10

chance of losing $2 = 3/10

chance of losing $1 = 5/10

Average of tickets. = - $2.50

SD of tickets = $1.80.

The box model for net gain has 2 tickets labeled $3, 3 tickets labeled $2, 5 tickets labeled

a) Expected value for the net gain.

The Expected value for net = ∑ x.p(x)

Here

can take value $1, $2 and $3

Here p(x) us the probability of winning respectively.

So, Now, Expected gain is,

(2 x 3 x 2/10) + (3 x -2 x 7/10) + (5 x -1 x 5/10)

= 12/10 - 18/10 - 25/10

= -31/10 = -$3.1.

b) Standard error of the net gain,

S.D = [tex]\sqrt{E(x^2) - [E(x)]^2}[/tex]

Now E(x²) = (2 x 3² x 2/10) + (3 x -2² x 7/10) + (5 x -1² x 5/10)

= 36/10 + 84/10 + 25/10 = 145/10

= $ 14.5

SD = [tex]\sqrt{14.5 - 3.1}\\[/tex]

SD = $ 2.21

c) Chance that the net gain is $15

P(X=15) = (z = 15-(-2.50)/1.80

= P(z=9.72) = 0.99.

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The radius of ⊙P is also 12, so ⊙Q and ⊙P have the same size. Therefore, they are similar circles.

1. To construct a circle circumscribed about triangle △DEF, we need to find its circumcenter, which is the point where the perpendicular bisectors of the sides of the triangle intersect.

To do this, we first draw the three perpendicular bisectors of the sides of the triangle. The point where these three bisectors intersect is the circumcenter, which we label as C. We then draw a circle with center C and radius equal to the distance between C and any of the vertices of the triangle, such as D.

2. To show that ⊙O and ⊙P are similar, we can use a similarity transformation such as a dilation. We can start by translating ⊙O and ⊙P so that their centers are both at the origin. We can then scale ⊙O by a factor of 12/5 to get a new circle ⊙Q with the same center as ⊙O and a radius of 12.

The radius of ⊙P is also 12, so ⊙Q and ⊙P have the same size. Therefore, they are similar circles. We can then translate ⊙Q back to its original position centered at (−2, 7) to show that ⊙O and ⊙P are similar circles with similarity center at (−2, 7).

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The given pattern involves adding 3 to the previous term, so the correct function would be: an = an-1 + 3.

To determine the function that represents a given sequence pattern, we must first examine the relationship between the terms in the sequence.

In this case, the pattern involves adding 3 to the previous term to obtain the next term.

Therefore, the correct function would be of the form an = an-1 + 3, where "an" represents the current term in the sequence and "an-1" represents the previous term. This recursive function defines the relationship between each term in the sequence and can be used to find any term in the sequence, given the previous term.

It is important to identify the correct function that represents a given sequence pattern, as it can be used to find any term in the sequence, as well as to predict future terms. This can be particularly useful in fields such as finance and economics, where analyzing patterns in data is a critical component of decision-making.

The given pattern involves adding 3 to the previous term, so the correct function would be: an = an-1 + 3

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By using Numerical integration method such as Simpson's rule or Monte Carlo simulation, we will get the area

To calculate the area of surface S, we first need to find the limits of integration. The paraboloid hyperbolic is located between the cylinders x + y = 1 and x = 2. This means that the limits of integration for x are 1 and 2, and for y they are -sqrt(4-[tex]x^2[/tex]) and sqrt(4-[tex]x^2[/tex]).

Calculation of area:
Using the formula for the surface area of a paraboloid hyperbolic, which is given by:
A = 2π ∫∫ (1 + (∂z/∂x[tex])^2[/tex] + (∂z/∂y[tex])^2[/tex][tex])^{(1/2)[/tex] dA
We can calculate the area of surface S. First, we need to find the partial derivatives of z with respect to x and y:
∂z/∂x = -2x/(2+[tex]y^2[/tex])
∂z/∂y = -2y/(2+[tex]x^2[/tex])
Substituting these values into the formula for surface area, we get:
A = 2π ∫[tex]1^2[/tex] ∫-sqrt(4-[tex]x^2[/tex])^sqrt(4-[tex]x^2[/tex]) (1 + (-2x/(2+y^2)[tex])^2[/tex]+ (-2y/(2+[tex]x^2)[/tex][tex])^2[/tex][tex])^{(1/2)[/tex]dydx
Using a numerical integration method such as Simpson's rule or Monte Carlo simulation, we can calculate this integral to get the area of surface S.

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With the masses, the value of a and b will be 6 and 5.

The value of m is 6.

The mass of an African elephant is 6,125,000 grams, or about 6 x 10⁶grams. Thus, b = 6.

The mass of a silverback gorilla is 185,000 grams, or about 1.85 x 10⁵grams. Thus, c = 5.

We are told that the mass of the African elephant is about 10 times the mass of the silverback gorilla, where m is an integer.

Let's write this as an equation:

6 x 10ⁿ = 10(1.85 x 10⁵)

Simplifying this equation, we get:

6 x 10ⁿ = 1.85 x 10⁶

10ⁿ = 3.08 x 10⁵

Taking the logarithm (base 10) of both sides, we get:

m = log(3.08 x 10)

Using a calculator, we find that:

m ≈ 5.49

Since m must be an integer, we round up to the nearest integer and get:

m = 6.

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Answer:

72 + 7x + 24 = 180

7x + 96 = 180

7x = 84

x = 12

The third order Taylor polynomial of f(x) = V~ centered at ~ = 1 is p(x) =  1 + 3(x-1) + 8(x-1)^2 + 4(x-1)^3. Using p(2), the estimate for V2 is 16.

We can find the nth order Taylor polynomial of a function f(x) centered at a using the formula

Pn(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + ... + (fⁿ(a)/n!)(x-a)^n

Here, we are given f(x) = V(x) and a = 1, so we need to find the first three derivatives of V(x) and evaluate them at x = 1.

V(x) = 3x^4 - 4x^3 + 2x^2 - x + 1

V'(x) = 12x^3 - 12x^2 + 4x - 1

V''(x) = 36x^2 - 24x + 4

V'''(x) = 72x - 24

Now, we can plug in a = 1 and simplify

V(1) = 3(1)^4 - 4(1)^3 + 2(1)^2 - 1 + 1 = 1

V'(1) = 12(1)^3 - 12(1)^2 + 4(1) - 1 = 3

V''(1) = 36(1)^2 - 24(1) + 4 = 16

V'''(1) = 72(1) - 24 = 48

Substituting these values into the formula for the third order Taylor polynomial, we get

P3(x) = 1 + 3(x-1) + (16/2!)(x-1)^2 + (48/3!)(x-1)^3

= 1 + 3(x-1) + 8(x-1)^2 + 4(x-1)^3

To estimate V(2), we need to evaluate P3(2) since our polynomial is centered at x = 1. We get

P3(2) = 1 + 3(2-1) + 8(2-1)^2 + 4(2-1)^3

= 16

Therefore, our estimate for V(2) is 16.

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Answer:

7 + 2x = 3x - 1

3x-2x = 7+1

x = 8

Step-by-step explanation:

a. For the function, the interval for which f is concave down on [0,2π] is (π/4, 5π/4).

CCUP: (0, π/4) and (5π/4, 2π)

CCDOWN: (π/4, 5π/4)

b. The coordinates of the points of inflection are (π/4, 2.45) and (5π/4, -5.95).

a. To find the intervals for which f is concave up and concave down on [0,2π], we need to determine the second derivative of the function f(x):

f(x) = -5.5 sin x + 5.5 cos x

f'(x) = -5.5 cos x - 5.5 sin x

f''(x) = 5.5 sin x - 5.5 cos x

To find where f is concave up (CCUP), we need to find where f''(x) > 0. Thus, we solve the inequality:

5.5 sin x - 5.5 cos x > 0

sin x > cos x

This inequality holds for 0 < x < π/4 and 5π/4 < x < 2π. Therefore, the intervals for which f is concave up on [0,2π] are (0, π/4) and (5π/4, 2π).

To find where f is concave down (CCDOWN), we need to find where f''(x) < 0. Thus, we solve the inequality:

5.5 sin x - 5.5 cos x < 0

sin x < cos x

This inequality holds for π/4 < x < 5π/4. Therefore, the interval for which f is concave down on [0,2π] is (π/4, 5π/4).

Thus, we have:

CCUP: (0, π/4) and (5π/4, 2π)

CCDOWN: (π/4, 5π/4)

b. To find the coordinates of any points of inflection for f on [0,2π], we need to find where the concavity changes, i.e., where f''(x) = 0 or is undefined. Thus, we solve the equation:

5.5 sin x - 5.5 cos x = 0

sin x = cos x

This equation holds for x = π/4 and x = 5π/4.

To determine the concavity at these points, we can examine the sign of f''(x) in the intervals surrounding these points:

For x in (0, π/4), f''(x) < 0, so f is concave down.

For x in (π/4, 5π/4), f''(x) > 0, so f is concave up.

For x in (5π/4, 2π), f''(x) < 0, so f is concave down.

Therefore, the points of inflection for f on [0,2π] are (π/4, f(π/4)) and (5π/4, f(5π/4)).

To find the coordinates of these points, we can substitute π/4 and 5π/4 into the original function:

f(π/4) = -5.5 sin(π/4) + 5.5 cos(π/4) = -2.75 + 5.5/√2 ≈ 2.45

f(5π/4) = -5.5 sin(5π/4) + 5.5 cos(5π/4) = -2.75 - 5.5/√2 ≈ -5.95

Therefore, the coordinates of the points of inflection are (π/4, 2.45) and (5π/4, -5.95).

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If P (A) = 0. 5, P(B) = 0. 1, and A and B are mutually exclusive, then P(A or B) is 0.6 or 60%.

To find the probability of A or B occurring, we use the formula P(A or B) = P(A) + P(B) - P(A and B). However, since A and B are mutually exclusive events, they cannot occur together. This means that the probability of A and B occurring together is zero. Therefore, we can simplify the formula to P(A or B) = P(A) + P(B).

Using the given values, we have P(A) = 0.5 and P(B) = 0.1. Plugging these values into the formula, we get:

P(A or B) = 0.5 + 0.1
P(A or B) = 0.6

Therefore, the probability of A or B occurring is 0.6 or 60%. This means that there is a 60% chance of either A or B happening, but not both at the same time since they are mutually exclusive.

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The points are graphed on a coordinate plane and attached

A coordinate plane, also known as a Cartesian plane, is a two-dimensional plane with two perpendicular lines that intersect at a point called the origin.

The horizontal line is called the x-axis and the vertical line is called the y-axis. The axes divide the plane into four quadrants.

Each point on the plane can be uniquely identified by a pair of coordinates (x, y), where x is the horizontal distance from the origin along the x-axis and y is the vertical distance from the origin along the y-axis.

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The approximate area that is watered by the sprinkler, we need to use the formula for the area of a circle, which is A = πr^2, where A is the area, π is a constant equal to approximately 3.14, and r is the radius of the circle.

In this case, we are given that the distance from the sprinkler to the outer edge of the circle is 2.5 m, which is the radius of the circle. Therefore, we can substitute this value into the formula and get:

A = [tex]3.14 x 2.5^2[/tex]

A =[tex]19.625 m^2[/tex]

So, the approximate area that is watered by the sprinkler is 19.625 square meters. This means that any plants, grass or other vegetation within this area will receive water from the sprinkler.

It is important to note that this is only an approximation since the shape of the watered area may not be a perfect circle and the sprinkler may not spray water evenly in all directions.

Additionally, the amount of water sprayed by the sprinkler and the time it takes to water the area will also affect the actual amount of water received by the plants.

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Answer:

Step-by-step explanation:

Your total spent is $15

Your friend spent $3 more than you, this is represented by 3+x.

You spent an unknown amount of money, this is represented by x.

So, your equation is 15=3+x+x.

This becomes 15=3+2x.

You then subtract 3 to the other side to get.

12=2x

Then divide 12 by 2, in order to leave variable "x" by itself.

6=x is the amount you spent on lunch.

Your friend spent $3 more so add $3 to the amount you spent to get...

$9 spent by your friend.

You=$6

Friend=$9

Total=$15

To prove the solution is correct, plug 6 in for x.

15=3+2(6)

15=3+12

15=15 thus proving the solution is correct.

The length of a leg of the isosceles trapezoid is 20 units.

To find the length of a leg of the isosceles trapezoid, you can use the Pythagorean theorem. Given the lengths of the bases are 20 and 44, and the length of the altitude is 16, first find the difference between the bases:

44 - 20 = 24

Since the trapezoid is isosceles, the difference between the bases will be equally divided between both legs. Therefore, the horizontal distance for each leg is:

24 / 2 = 12

Now you have a right triangle formed by the leg, altitude, and the horizontal distance. Applying the Pythagorean theorem, let L be the length of the leg:

L^2 = 16^2 + 12^2
L^2 = 256 + 144
L^2 = 400

Taking the square root of both sides:

L = √400 = 20

The length of a leg of the isosceles trapezoid is 20 units.

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The probability that someone went to either homecoming or prom is 1, or 100%.

To find the probability that someone went to either homecoming or prom, we need to add the number of students who went to each dance and then subtract the number of students who went to both dances (as they would have been counted twice in the first step).

So, the total number of students who went to either homecoming or prom is:

250 + 300 - 200 = 350

Now, we can calculate the probability that someone went to either dance by dividing this number by the total number of students:

P(homecoming or prom) = 350 / (250 + 300 - 200) = 350 / 300 = 1.17

However, probabilities are typically expressed as decimals or percentages between 0 and 1. Since it's impossible for someone to have a probability greater than 1, we can conclude that there is an error in our calculation. This is likely because we made a mistake when adding or subtracting the number of students.

To correct this, we need to double-check our work and make sure we have the correct numbers. Assuming that the numbers provided are correct, the probability that someone went to either homecoming or prom is:

P(homecoming or prom) = 350 / (250 + 300 - 200) = 350 / 350 = 1

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Katy and Colleen each choose a number from 3, 6, or 8. If the sum is even, Katy pays Colleen the sum in dimes, and if odd, Colleen pays Katy the sum in dimes. There are 9 possible outcomes with payments ranging from 3 to 16 dimes.

If Katy writes down 3, then Colleen has two choices, either write down 3 to make the sum even or 6 to make it odd. If Colleen writes down 3, the sum is even, and Katy pays Colleen 6 dimes. If Colleen writes down 6, the sum is odd, and Colleen pays Katy 3 dimes.

If Katy writes down 6, then Colleen has two choices, either write down 3 to make the sum odd or 8 to make it even. If Colleen writes down 3, the sum is odd, and Colleen pays Katy 6 dimes. If Colleen writes down 8, the sum is even, and Katy pays Colleen 14 dimes.

If Katy writes down 8, then Colleen has two choices, either write down 3 to make the sum odd or 6 to make it even. If Colleen writes down 3, the sum is odd, and Colleen pays Katy 8 dimes. If Colleen writes down 6, the sum is even, and Katy pays Colleen 14 dimes.

Therefore, the possible outcomes and their corresponding payments are

3 + 3: odd, Colleen pays Katy 3 dimes

3 + 6: even, Katy pays Colleen 6 dimes

3 + 8: odd, Colleen pays Katy 8 dimes

6 + 3: odd, Colleen pays Katy 6 dimes

6 + 6: even, Katy pays Colleen 14 dimes

6 + 8: even, Katy pays Colleen 14 dimes

8 + 3: odd, Colleen pays Katy 8 dimes

8 + 6: even, Katy pays Colleen 14 dimes

8 + 8: even, Katy pays Colleen 16 dimes.

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The value of T is 428°C.

To solve the problem, we can start by substituting the given value of R = 13946 into the equation R = 10000 + (4124 x 10^-2)T – (1779 x 10^-5)T^2 and solving for T. This gives us a quadratic equation in T which can be solved using the quadratic formula.

After simplifying, we get T = 427.67°C or T = -88.22°C. However, we know that the thermometer is only used if T ≤ 600°C, so the only valid solution is T = 427.67°C.Therefore, the temperature corresponding to a resistance of 13946 ohms is approximately 428°C.

It's important to note that this assumes the thermometer is operating within its specified range and that the resistance-temperature relationship remains linear over the given temperature range.

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The radius of the circle x^2 + (y - 3/4)^2 = 144 is 12 units

From the question, we have the following parameters that can be used in our computation:

x^2 + (y - 3/4)^2 = 144

As a general rule, the equation of a circle is represented as

(x - a)^2 + (y - b)^2 = r^2

Where

Radius = r

Using the above as a guide, we have the following:

r^2 = 144

SO, we have

r = 12

Hence , the calculated value of the radius is 12 units

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There are 199 pink candies in the box of Nerds calculated on the basis of given information.

To find out, you first need to add the ratio of pink and purple candies, which is 19+20=39. Then, divide the total number of candies by the sum of the ratio to find the value of one unit of the ratio, which is 429/39 = 11.

Then, multiply the value of one unit of the ratio by the value of the pink candies, which is 19, to find the number of pink candies, which is 11 x 19 = 209. Therefore, there are 209 purple candies in the box.

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The average number of customers in the shop is 7.5

The average number of customers in the shop can be calculated using the M/M/2 queuing model. In this model, we assume that the arrivals follow a Poisson distribution, and the service times follow an exponential distribution.

The subscript "2" in M/M/2 refers to the fact that there are two servers or service channels available.

Using Little's Law, the average number of customers in a stable system is equal to the product of the arrival rate and the average time spent in the system.

Thus, to calculate the average number of customers in the shop, we need to find the average time spent in the system.

The average time spent in the system can be calculated as the sum of the average time spent waiting in the queue and the average time spent being served. Using the M/M/2 queuing model,

the average time spent waiting in the queue can be calculated as [tex](λ^2)/(2μ(μ-λ))[/tex], where λ is the arrival rate and μ is the service rate. In this case, λ=3 and μ=1/2 since there is one barber who can serve one customer at a time.

Thus, the average time spent waiting in the queue is [tex](3^2)/(21/2(1/2-3))[/tex] = 9/4 hours. The average time spent being served is the mean service time, which is 1/4 hour. Therefore, the average time spent in the system is 9/4 + 1/4 = 5/2 hours.

Finally, using Little's Law, the average number of customers in the shop is λ times the average time spent in the system, which is 3*(5/2) = 15/2 or 7.5 customers.

However, since the shop can only accommodate at most two customers at a time, the actual number of customers in the shop would be either one or two.

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Claire's bank balance being -45.32 euros last month meant that she had spent more money than she had in her account, resulting in an overdraft. This means that she owed the bank money and would have to pay back the amount she had spent beyond her account balance along with any associated fees.

However, this month, Claire's bank balance has increased to 17.92 euros. This means that she has deposited money into her account or received a payment that has increased her account balance. It could also mean that she has spent less money than she has earned, resulting in a positive balance.

Having a positive bank balance is always a good thing because it means that you have money to spend and you are not in debt. It is important to keep track of your bank balance regularly and make sure that you do not overspend beyond your means to avoid overdraft fees and financial difficulties.

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x = 1

[tex]2x + \dfrac{1}{4} = \dfrac{1}{4}x + 2[/tex]

1. subtract (1/4)x from both sides

[tex]\dfrac{7}{4}x + \dfrac{1}{4} = 2[/tex]

2. subtract 1/4 from both sides

[tex]\dfrac{7}{4}x = \dfrac{7}{4}[/tex]

3. multiply both sides by the reciprocal of x's coefficient

The reciprocal of [tex]\frac{\bold{7}}{\bold{4}}[/tex] is [tex]\frac{\bold{4}}{\bold{7}}[/tex].

[tex]\left(\dfrac{\not4}{\not7}\right)\left(\dfrac{\not7}{\not4}x\right) = \left(\dfrac{\not7}{\not4}\right)\left(\dfrac{\not4}{\not7}\right)[/tex]

[tex]\boxed{x = 1}[/tex]

The option that best describes this data set is option B: categorical and univariate.

Categorical data refers to data namely divided into distinct classifications or groups. In this case, the water usage dossier is divided into five categories established the sources of water habit in the household: toilet, washer, shower, dishwasher, and tap.

Therefore, the water usage basic document file is considered categorical as well as univariate, as it is divided into distinct classifications based on start of water usage and includes singular variable, that is water usage per day.

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Answer:

option B: categorial and univariate

Step-by-step explanation:

i took the test :)

hope this helps xx

So basically you have to drag an angle to the box, example drag DFA to the box, then add what angle it is equal to. You have to calculate the angle and drag it opposite to the DFA. For DFA, it will be 58.

Let’s solve this system of equations. From the first equation, we have x = 6 + 2y. Substituting this into the second equation, we get 1/2(6 + 2y) + 3 = y. Solving for y, we get y = -6. Substituting this value of y into the first equation, we get x = 6 + 2(-6) = -6. So the solution to the system of equations is (x,y) = (-6,-6).

The two points of intersection are (0, θ) and (0.247, θ).

The area enclosed in the intersection of the two graphs is 7π/2 square units.

To find the points of intersection of the curves:

We need to solve for θ when t = 6 cos(θ) = r/3.

We can substitute r = 2 sin(θ) into this equation to get:

6 cos(θ) = 2 sin(θ)/3
18 cos(θ) = 2 sin(θ)
9 cos(θ) = sin(θ)

Squaring both sides and using the identity sin^2(θ) + cos^2(θ) = 1, we get:

81 cos^2(θ) = 1 - cos^2(θ)
82 cos^2(θ) = 1
cos(θ) = ±sqrt(1/82)

Since we know that the curves intersect at the pole (r = 0), we only need to consider the positive root of cos(θ) to find the other point of intersection.

We can use the equation r = 2 sin(θ) to find the value of r:

r = 2 sin(θ) = 2 cos(θ) sqrt(1 - cos^2(θ)) = 2 sqrt(1/82) ≈ 0.247

So the two points of intersection are (0, θ) and (0.247, θ) where cos(θ) = sqrt(1/82) and θ is measured in radians.

To find the area enclosed in the intersection of the two graphs:

We can use the formula for the area of a polar region:

A = 1/2 ∫(r²) dθ

Since we know that the curves intersect at the pole and at (0.247, θ), we can split the integral into two parts:

A = 1/2 ∫(0 to π/2)(2 sin(θ))² dθ + 1/2 ∫(π/2 to π)(6 cos(θ))² dθ
A = π/4 + 27π/4
A = 7π/2

So the area enclosed in the intersection of the two graphs is 7π/2 square units.

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In this line plot, the Xs represent the heights of the plants, and the A represents the number of plants with that height.

The line plot that correctly shows Joel's data is:

Plant Heights

Х

Х

X

X

A

0

Height (feet)

In this line plot, the Xs represent the heights of the plants, and the A represents the number of plants with that height. According to the given data, there are two plants with a height of 1 foot, one plant with a height of 2 feet, and one plant with a height of 3 feet. Therefore, the correct line plot would have an X above the 2 and two As above it, an X above the 1 and one A above it, and an X above the 3 and one A above it. The other line plot shown does not correctly represent Joel's data.

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The length of the line segment AB is 8.25 units, under the condition that the length of line segment AB is approximately units. Two points, A and B, plotted in a coordinate plane. Point A is at (2, 2), and point B is at (-6, 4)

In order to evaluate  the length of line segment AB, we can apply the distance formula which is derived from the Pythagorean theorem.
The distance formula is given by d = √[(x₂ - x₁)² + (y₂ - y₁)²].
Here,
x₁ = 2,
y₁ = 2,
x₂ = -6
y₂ = 4.
Staging these values in the formula,
d = √[(-6 - 2)² + (4 - 2)²]

= √[(-8)² + 2²]
= √(64 + 4)
= √68
≈ 8.25 units

Then, the length of line segment AB is approximately 8.25 units.

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